3.8.89 \(\int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx\) [789]

Optimal. Leaf size=220 \[ \frac {\sqrt {2} (a+b) F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt {1+\sec (e+f x)}}-\frac {\sqrt {2} a F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt {1+\sec (e+f x)}} \]

[Out]

(a+b)*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sec(f*x+e))/(a+b),1/2-1/2*sec(f*x+e))*(a+b*sec(f*x+e))^m*2^(1/2)*tan(f*x+
e)/b/f/(((a+b*sec(f*x+e))/(a+b))^m)/(1+sec(f*x+e))^(1/2)-a*AppellF1(1/2,-m,1/2,3/2,b*(1-sec(f*x+e))/(a+b),1/2-
1/2*sec(f*x+e))*(a+b*sec(f*x+e))^m*2^(1/2)*tan(f*x+e)/b/f/(((a+b*sec(f*x+e))/(a+b))^m)/(1+sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3923, 3919, 144, 143} \begin {gather*} \frac {\sqrt {2} (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt {\sec (e+f x)+1}}-\frac {\sqrt {2} a \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt {\sec (e+f x)+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2*(a + b*Sec[e + f*x])^m,x]

[Out]

(Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)]*(a + b*
Sec[e + f*x])^m*Tan[e + f*x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m) - (Sqrt[2]*a*Appel
lF1[1/2, 1/2, -m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)]*(a + b*Sec[e + f*x])^m*Tan[e + f*
x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m)

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 3919

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Cot[e + f*x]/(f*Sqr
t[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x]]), Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f
*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*m]

Rule 3923

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[-a/b, Int[Csc[e +
f*x]*(a + b*Csc[e + f*x])^m, x], x] + Dist[1/b, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; Free
Q[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx &=\frac {\int \sec (e+f x) (a+b \sec (e+f x))^{1+m} \, dx}{b}-\frac {a \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx}{b}\\ &=-\frac {\tan (e+f x) \text {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}+\frac {(a \tan (e+f x)) \text {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {\left (a (a+b \sec (e+f x))^m \left (-\frac {a+b \sec (e+f x)}{-a-b}\right )^{-m} \tan (e+f x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}+\frac {\left ((-a-b) (a+b \sec (e+f x))^m \left (-\frac {a+b \sec (e+f x)}{-a-b}\right )^{-m} \tan (e+f x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (e+f x)\right )}{b f \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {\sqrt {2} (a+b) F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt {1+\sec (e+f x)}}-\frac {\sqrt {2} a F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt {1+\sec (e+f x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(5564\) vs. \(2(220)=440\).
time = 20.44, size = 5564, normalized size = 25.29 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[e + f*x]^2*(a + b*Sec[e + f*x])^m,x]

[Out]

Result too large to show

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \left (\sec ^{2}\left (f x +e \right )\right ) \left (a +b \sec \left (f x +e \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2*(a+b*sec(f*x+e))^m,x)

[Out]

int(sec(f*x+e)^2*(a+b*sec(f*x+e))^m,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(a+b*sec(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e) + a)^m*sec(f*x + e)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(a+b*sec(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e) + a)^m*sec(f*x + e)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sec {\left (e + f x \right )}\right )^{m} \sec ^{2}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2*(a+b*sec(f*x+e))**m,x)

[Out]

Integral((a + b*sec(e + f*x))**m*sec(e + f*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2*(a+b*sec(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e) + a)^m*sec(f*x + e)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^m}{{\cos \left (e+f\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x))^m/cos(e + f*x)^2,x)

[Out]

int((a + b/cos(e + f*x))^m/cos(e + f*x)^2, x)

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